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Energy
Introduction
Some of the most powerful tools in physics are based on conservation principles.
The idea behind a conservation principle is that there are some properties of systems
that don’t change, even though other things about the system may.
For instance, let’s say that I have a package of candy that contains exactly 50 pieces. If
I take those pieces of candy out of the package and put them on top of a table, I still
have 50 pieces. If I lay them end-to-end or arrange them into a rectangle, I still have 50
pieces. No matter how many different ways I arrange them; I still have 50 pieces. They
may look different in each case, but the total number stays the same. In this example, I
could say that the number of pieces of candy is conserved.
Energy is another example of a conserved property of a system. It’s hard to come up
with a meaningful definition of energy. It’s a basic property of the universe, like time
and space, so it’s very hard to define. It’s a lot easier to visualize a piece of candy than
a piece of energy. However, it is possible to mathematically describe the various forms
of energy. Having done that, it has consistently proven true that if you add up all the
types and amounts of energy within a closed system the total amount of energy does
not change.
To work with this definition it’s important to understand the idea of a closed system. It is
true that I could change the number of pieces of candy on the table by eating some of
them, dropping a piece on the floor or opening another package and spilling some of
that new candy onto the table. We have to account for any candy that’s been added to
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or taken away from the amount that we started with or we’ll see that the number has
changed and our conservation principle will seem to have been violated.
The same thing is true for energy. The amount of energy in a closed system stays
constant. But that means that if we add energy to the system or take energy away from
a system, we have to account for it. Unless we do that successfully, it will appear that
the Conservation of Energy principle has been violated. One way that we can move
energy into or out of a system is called Work. Work has a very specific mathematical
definition in physics and it represents the movement of mechanical energy into or out of
a system. If work is the only means to move energy into or out of our system than it will
be true that
Initial Amount of Energy + Work = Final Amount of Energy
Or
E0 + W = Ef
The forms that energy takes can vary quite widely. Some of these forms include
gravitational, electrical, chemical, kinetic, magnetic, elastic and nuclear. These are just
some of the forms that energy may take, but there are many more. In this chapter, we’ll
be discussing several mechanical forms of energy, kinetic energy, gravitational
potential energy and elastic potential energy, along with the concept of work.
Work
Work is defined as the product of the force applied to an object and the distance that the
object moves in the direction of that force. The mathematical description of that
definition is:
Work = Force x Distance parallel
Or
W = Fd parallel
It is important to note that work is proportional to the product of the force and the
distance that the object moves parallel to that force. That means that if the object
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moves in the direction that I am pushing or pulling it, then I am doing work. If it does not
move, or if it moves perpendicular to the direction that I am pushing or pulling it, I am
not doing any work.
This can be confusing because the use of the word “work” in English is similar to but not
the same as its use in physics. For instance if someone were to pay me to hold a heavy
box up in the air while they move a table to sweep underneath it, I would say that I am
doing work. But I would not be doing work according to the physics definition of the
term. That is because the box is not moving in the direction of the force that I am
applying. I am applying a force upwards but the box is stationary. Since d parallel is equal
to zero, so is the amount of work, W.
The same thing applies if I were to put that heavy box on a perfectly frictionless cart and
push it to the side of the room at a constant velocity. Since the velocity is constant, the
acceleration is zero. If there’s no friction to overcome, then the force I need to apply
(once I’ve gotten it moving) is zero. No force… no work.
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The last way that I can do no work, according to the physics definition of work, is if I
were to carry that box across the room at a constant velocity and put it on a shelf at the
same height. Once again, I don’t need to apply a horizontal force to keep moving at a
constant velocity so there are no forces in the horizontal direction. I am applying a force
in the vertical direction, to keep the box from falling to the ground. But the box is not
moving in the vertical direction, it’s moving in the horizontal direction. So in the
horizontal direction, the force is equal to zero and in the vertical direction d parallel equals
zero. The result is that W = 0 in both cases.
While our definition of work may not always seem to relate to our experience, it turns
out to be a very useful tool in developing a theory of energy. In fact, the three forms of
energy that we will be discussing in this chapter all become clear through thinking about
them with respect to work.
Units of Energy
The unit of energy can be derived from the basic equation of work.
W = F x d parallel
The SI units of force are Newtons (N) and of distance are meters (m). Therefore, the
units of energy are Newton-meters (N-m). Out of respect for James Prescott Joule
(1818-1889), a key formulator of the concept of energy, this is also referred to as a
Joule (J).
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J = N-m
J = (kg–m / s2) - m
J = N-m = kg-m2 / s2
Example 1: A constant force of 45 N is applied to a mass on a frictionless surface. The
force is applied in the same direction as the motion of the object. How much work does
that force do over a distance of 6.0m?
Since the force and the distance that the object moves are parallel to one another the
work done by the force will simply be the product of the two.
W = Fdparallel
= 45 N x 6m
= 270 N-m
= 270 J
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Example 2: A force of 45N keeps an object moving in circular motion at a constant
speed on a horizontal frictionless surface. The circumference of the circle is 6.0m. How
much work does that force do during one rotation?
The force needed to keep an object moving in a circle at a constant speed on a
horizontal frictionless surface is directed towards the center of the circle. However, the
velocity of the object is always tangent to the circle. Therefore F and d are always
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perpendicular to one another. As a result, both d parallel and the work done by the force
are equal to zero.
Gravitation Potential Energy
Imagine lifting a box off the floor of your room and putting it on a shelf. The shelf is at a
height “h” and the box has a mass “m”. Lifting the box off the floor requires you to
supply a force at least equal to its weight, “mg”. That means that you have to do some
work, since you are lifting the box in the same direction that it moves, straight up. (In
fact, at the beginning you had to supply a little more than that amount of force to get it
moving up and at the end a little less than that to get it to slow down to a stop, but on
average the force applied would exactly equal mg.)
W = F x d parallel
W = (mg) h
W = mgh
But our definition of conservation of energy tells us that
E0 + W = Ef
That means that the work you just did must have added energy to the system in the
amount of
W = E f - E0
Since the work that you did was equal to mgh that means that the energy of the system
must have been increased by that amount
Ef - E0 = mgh
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This extra energy must be stored in some form. This leads us to our first type of
energy, Gravitation Potential Energy (GPE). The term GPE reflects the fact that this
energy is due to the change of height of a mass located in the earth’s gravitational field.
The Gravitational Potential Energy, GPE, of a system is given by:
GPE = mgh
Example 3: Determine the GPE (relative to the floor) of a 50 kg box located 12m above
the floor.
GPE = mgh
= (50kg)(9.8 m/s2)(12m)
= 5880 kg-m2/s2
= 5900 J
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Example 4: A 52 kg man walks down a 4.2m tall flight of stairs. How much work did he
do?
E0 + W = Ef
W = Ef - E0
= 0 - GPE0
= - mgh0
= -(52kg)(9.8 m/s2)(4.2m)
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= - 2140 kg-m2/s2
W = -2100 J
Kinetic Energy
Remember the box that you put up on the shelf a few pages ago. Well, simply knocking
that box off the shelf will allow us to derive our second form of energy. The box will fall
towards the floor with an acceleration equal to g. We can determine its velocity just
before impact by using our third kinematics equation.
v2 = v02 + 2ad
v2 = 0 + 2gh
v2 = 2gh
gh = v2 / 2
The box is now just above the floor and moving with the velocity Indicated above. No
energy moved into or out of the system during this process. That means the energy is
the same as it was when the box was on the shelf, mgh. Therefore
E0 + W = Ef
E0 + 0 = Ef
mgh = Ef
gh = Ef / m
We can now set this equal to the expression for “gh” that we derived above
v2/2= Ef / m
Or
Ef = ½ mv2
Just before the mass strikes the ground its GPE (relative to the ground) must equal zero
since its height is zero. The box still has the same amount of energy but it’s in a new
form since it is no longer in the form of GPE. In this case, the energy is stored in the
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form of a mass moving with a velocity. That form of mechanical energy is called Kinetic
Energy and is given by
KE = ½ mv2
In knocking the box off the shelf we started a process that converted its Gravitational
Potential Energy into Kinetic Energy. It went from being a motionless mass at a height
to a moving mass near the floor.
We can’t follow our box as it strikes the floor, as that would introduce a whole new set of
complicated energies. The reason it was so difficult to develop a good energy theory
was these very complexities. We would have to account for the energy that gets
transferred into the motion of the atoms that comprise the box and the floor (heat), the
crashing sound of the box hitting the floor (sound energy) and the permanent
deformation of the box and the floor.
It is possible to account for these in a more detailed study, but that is not necessary for
what we need to do in this course. It is worth pointing out however, that the
experiments of Joule proved decisive in putting the energy theory on solid ground.
Joule was able to show that thermal energy was able to account for the “missing
energy” in many examples like the one that we have been discussing. By doing this,
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the way was cleared to develop a complete and powerful theory. For this reason, the
unit of energy used in the SI system is called the Joule.
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Example 5: Determine the KE of a 3.0 kg ball which has a velocity of 2.1 m/s.
KE = ½ mv2
= ½ (3.0kg)(2.1m/s)2
= (1.5 kg)(4.41m2/s2)
= 6.615 kg-m2/s2
= 6.6 J
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Example 6: Use conservation of energy to determine how high a ball will go if it leaves
the ground with a velocity of 24 m/s.
E0 + W = Ef
but W = 0 so
E0 = E f
the energy is either GPE or KE so
KE0 + GPE0 = KEf + GPEf
then substitute in the formulas for each
½
mv02
½
mv02
½
v02
+ mgh0 = ½
m(vf)2+
mghf
but h0 and vf are both = 0 so
= mghf
divide both sides by m to cancel it out
= ghf
double both sides
v02 = ghf
divide both sides by g
hf = v02 / g
substitute in the given values
hf = (24 m/s)2 / (9.8 m/s2)
hf = (576 m2/s2) / (9.8 m/s2)
hf = 59 m
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It’s important to note that the height to which an object will rise does not depend on its
mass. It only depends on its initial velocity.
Elastic Potential Energy
The final form of energy that we will be discussing in this chapter is Elastic Potential
Energy. It represents the energy that can be stored in a spring. First, we must
understand the force that is required to compress or stretch springs. This was first
explained by Robert Hooke and is therefore referred to as Hooke’s law.
Hooke observed that it takes very little force to stretch a spring a very small amount.
However, the further the spring it stretched, the harder it is to stretch it further. The
force needed increases in proportion to the amount that it has already been stretched.
The same was observed to be true when compressing a spring. This can be stated
mathematically as
Fspring = - k x
In this equation, k represents the spring constant (a characteristic of the individual
spring), and x represents the distance the spring is stretched or compressed from its
natural length. The negative sign tell us that the force that the spring exerts is back
towards its equilibrium length, its length when it is not being stretched or compresses.
Therefore, if the spring constant for a particular spring were 100 N/m, I would need to
exert a force of 100 Newtons to stretch, or compress, it by a length of 1m. If I were to
exert a force of 50N, it would stretch 1/2 m. A force of 10 N would stretch, or compress,
it by a distance of 1/10 m. This is shown graphically below.
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It is worthwhile to compare the above graph to a graph of you lifting the box onto the
shelf in our discussion of gravitational potential energy. In that case, the force was
constant, mg, regardless of the distance that the box was raised. The graph of that is
shown below. Also shown on that graph is the fact that the area of the rectangle that is
formed by the force, mg, on the y-axis and the distance, h, on the x-axis, is equal to
mgh. This is equal to the work needed to lift the box to the height of the shelf (and also
to the gravitational potential energy it had upon attaining that height). Thus the work
needed to lift the box is equal to the area under the Force versus Distance curve.
Similarly, in the case of a spring, the work needed to stretch or compress it will be equal
to the area under its force versus distance curve. However, as can be seen below, in
this case the shape that is formed is a triangle with a height equal to kx (equal and
opposite to the force exerted by the spring, -kx) and a base equal to x. The area of a
triangle is given by ½ base times height. Therefore the work needed to compress or
stretch a spring a distance x is given by
W = Area under the F vs. d curve
That curve forms the indicated triangle.
W = ½ (base) (height)
W = ½ (x) (kx)
W = ½ kx2
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The energy imparted to the spring by this work must be stored in the Elastic Potential
Energy (EPE) of the spring. Therefore,
EPE = ½ kx2
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Example 7: Determine the energy stored in a spring whose spring constant, k, is 200 N
/ m and which is compressed by a distance of 0.11m from its equilibrium length
.
EPE = ½ kx2
= ½ (200 N/m)(.11m)2
= (100 N/m)(.0121 m2)
= 1.21 kg-m2/s2
= 1.2 J
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Example 8: A spring powered dart gun is used to shoot a dart straight upwards. The
mass of the dart is 40 gm. The spring constant of the spring in the dart gun is 500 N/m
and it is compressed a distance of 5.0 cm before being fired. Determine its velocity
upon leaving the gun and the maximum height it attains (measured from the its initial
location with the spring no longer compressed).
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For the first part of the problem we need to determine the dart’s velocity just at the
moment it leaves the gun. At that point, all the EPE of the spring will have converted to
KPE and GPE.
E0 + W = Ef
but W = 0 so
E0 = E f
the energy is either GPE or KE so
EPE0 = KEf + GPEf
then substitute in the formulas for each
½ kx2 = ½ mvf2+ mghf
subtract mghf from both sides
½ mvf2 =½ kx2 - mghf
divide both sides by m
½ vf2 = ½ (k/m)x2 - ghf
double both sides
vf2 = (k/m)x2 - ghf
substitute in the given values
vf2 = (500N/m)/.04kg))(.05m)2 - (9.8 m/s2)(.05m)
vf2 = 31 m2/s2- .49 m2/s2
vf = 5.5 m/s
For the second part of the problem we need to use 5.5 m/s as the initial velocity and
then determine from that how high it goes. We can use the work that we did in Example
6 to simplify the problem.
hf = v02 / g
hf = (5.5 m/s)2 / (9.8 m/s2)
hf = (30.2 m2/s2) / (9.8 m/s2)
hf = 3.1 m
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Power
It is often important to know not only if there is enough energy available to perform a
task but also how much time will be required. The concept of power allows us to
answer these questions.
Power is defined as the rate that work is done. This is expressed mathematically in
several forms. The most fundamental form is
Power = Work per unit time
P=W/t
Since W = Fd parallel this can also be expressed
P = (Fd parallel) / t
Regrouping this becomes
P = F(d parallel / t)
Since v = d/t
P = Fv parallel
So power can be defined as the product of the force applied and the velocity of the
object parallel to that force. A third useful expression for power can be derived from our
original statement of the conservation of energy principle.
P=W/t
Since W = Ef - E0
P = (Ef - E0) / t
So the power absorbed by a system can be thought of as the rate at which the energy in
the system is changing.
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Units of Power
The SI unit for power can be derived from its basic equation.
P=W/t
The SI unit of energy is the Joule (J) and of time is the second (s). Therefore, the SI
unit for power is a Joule per second (J / s). This has been designated the Watt (W) out
of respect for James Watt (1736 - 1819), a pioneer in the development of the steam
engine.
W = J/s
W = (N-m)/s
W = ((kg-m/s2)(m)/s
W = J/s = kg-m2 / s3
Example 7: What is the minimum power needed to lift a 120 kg mass 25 m straight up
in 8.0 seconds?
P=W/t
but W = Fd parallel so
P = Fd parallel / t
But the force needed to lift a mass is mg so
P = mgd parallel / t
substitute in the given values
P = (120kg) (9.8 m/s2)(25m)/(8.0s)
P = 3675 kg-m2/s3
P = 3700 W
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Example 8: If a motor can supply 45000 W of power, with what velocity can it lift a 1200
kg elevator car filled with up to 8 110 kg people.
First we’ll need to calculate the total mass of the car loaded with the maximum number
of people.
mtotal = mcar + 8 (mperson)
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mtotal = 1200kg + 8 (110kg)
mtotal = 1200kg + 880kg
mtotal = 2080 kg
Then
P = Fvparallel
Recognizing that F = mg and v = vparallel
P = (mg)v
Dividing both sides by mg yields
v = P/(mg)
Substituting values yields
v= 45000W/((2080kg)(9.8
m/s2))
v = 2.21 (kg-m2/s3)/(kg-m/s2)
v = 2.2 m/s
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Simple Machines
Simple machines make it possible for us to accomplish what might otherwise be difficult
or impossible. They have been used throughout history. Initially they were hand
powered. As technology evolved it proved possible to use other sources of power such
as water, wind, steam, internal combustion engines and electric motors to drive these
machines. However, regardless of their power source, simple machines still serve the
same function of allowing the best match between the source of energy and the work
that needs to be done.
Simple machines include the inclined plane, screw, lever and pulley. All of them work
on the basis of the equation that
W = Fd parallel
In all these cases the force will be made parallel to the distance moved so we can
simplify this expression to
W = Fd
Machines cannot create energy. Rather, they allow us to use energy more effectively
by varying the force and distance involved while leaving the product of those two
factors, the work, unchanged.
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For instance, let’s say that you needed to lift a 100 kg (220 lb) piece of equipment from
the ground to the back of a truck, 1.0 m above the ground. That would be virtually
impossible for most people without the aid of a simple machine. An average person
could not lift 220 lbs an inch, let alone a meter. The required force, in SI units, would be
given by
E0 + W = Ef
We can take E0 = 0 so
W = Ef
and W = Fd to get
Fd = Ef
But Ef = GPE = mghf so
Fd = mghf
Solve for F by dividing both sides by d
F = mghf / d
Substituting values
F = (100kg)(9.8 m/s2)/1.0m
F = 980 N
However, if a 5.0 m long inclined plane (using low friction bearings) is extended from the
back of the truck to the ground, the piece of equipment could be pushed up the ramp.
The effect of doing this is to increase d from 1.0 m to 5.0 m. Let’s calculate the force
needed to push the equipment up the ramp (assuming no friction).
E0 + W = Ef
We can take E0 = 0 so
W = Ef
and W = Fd to get
Fd = Ef
But Ef = GPE = mghf so
Fd = mghf
Solve for F by dividing both sides by d
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F = mghf / d
Substitute in values
F = (100kg)(9.8 m/s2)/5.0m
F = 200 N
Since d is now 5 times larger and W stays the same (this is where the frictionless part of
the plane is important) the needed force becomes 1/5 as large. Instead of requiring
980N, the task can be accomplished with a force of 200N (44 lb). Even with the
addition of friction, the task would go from being impossible to relatively easy. The
tradeoff of having to push the box five times as far is well worth it if the box is sufficiently
heavy.
This is the same principle that underlies all simple machines. When the goal is to
reduce the amount of force needed, W is kept constant while d is increased. The result
is that the force is decreased by the same multiple that the distance is increased.
In some cases the goal is to increase the force. For instance when a lever is used to
move a heavy boulder. A lever consists of a rigid rod or board atop a fulcrum. By
making the distance from your hands to the fulcrum larger than the distance from the
fulcrum to the object, the force exerted on that object is magnified.
Simple machines can work in either direction. If you push on the longer side of the lever
you magnify the force and reduce the distance. However, you could also push on the
shorter side to reduce the force and increase the distance. In either case, in a perfectly
efficient simple machine, the product of the force and distance is the same. Therefore,
the best way to look at these problems is to use the equation for work on both sides and
recognize that the conservation of energy principle assures that they are equal. The
work that I put in on my side of the fulcrum results in work being done to the object on
the other side.
Wout = W in
Substitute in W = Fd (F and d are parallel) on both
sides of the equation
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Foutdout = Findin
Divide both sides by dout
Fout = Findin /dout
Regroup the terms
Fout = Fin (din /dout)
Since din is proportional to the length of the lever on my side of the fulcrum and dout is
proportional to the length of the lever on the other side of the fulcrum, this machine
magnifies the force by a factor equal to the ratio of those two lengths. The tradeoff is
that although I’ve increased the amount of force, I have decreased the distance I will
move the object by the same factor.
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Example 9: I need to lift a 500 kg (1100 lb boulder) a few cm off the ground. The
closest that I can get the fulcrum to the bottom of the boulder is 10 cm. How long will
the lever need to be so that I could move the boulder by pushing down with all my
weight, 500N?
Wout = Win
Substitute in W = Fd (F and d are parallel) on both
sides of the equation
Foutdout = Findin
Divide both sides by Fin
Foutdout / Fin = din
Or
din = Foutdout / Fin
Substituting in the values with Fout = mboulderg
din = (500kg)(9.8 m/s2)(.1m)/500N
din = 0.98 m
But the lever consists of both sides, din + dout
L = din + dout
L = .1m + .98m
L = 1.1m
Chapter Questions
1. A block is suspended from a string; does the gravitational force do any work on it?
2. What is the difference between work done by the gravitational force on descending
and ascending objects?
3. A woman climbs up stairs; does she do any work? Does she do any work standing
in an ascending elevator?
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4. What happens to an object’s velocity if there is work done by a friction force? Why?
5. An object is suspended from a spring and is at equilibrium; does the elastic force do
any work?
6. It is known that water applies some pressure on a container; does water do any
work in this case?
7. What kind of energy does a flying bullet have?
8. A stone is thrown vertically up. What kind of energy did the stone have initially?
What happens to this energy as the stone ascends?
9. A steel ball and an aluminum ball of equal volume are located at the same altitude.
Which ball has greater gravitational potential energy?
10. What happens to the gravitational potential energy of an object when it moves up?
When it moves down?
11. Is it possible for a static friction force to do mechanical work? Give an example?
12. Can kinetic energy ever be negative? Explain.
13. Describe the energy transformation that takes place when a small mass oscillates at
the end of a light string.
14. Describe the energy transformations that take place when a small mass oscillates at
the end of an elastic spring.
15. An elevator is lifted vertically upwards at a constant speed. Is the net work done on
the elevator negative, positive, or zero? Explain.
16. Can the net work done on an object during a displacement be negative? Explain.
Chapter Problems
Work
1. A car engine applies a force of 65000 N, how much work is done by the engine
as it pushed a car a distance of 75 m?
2. A force does 30000 J of work along a distance of 9.5 m. Find the applied force.
3. How high can a 40 N force move a load, when 395 J of work is done?
4. A boy pulls a sled at a constant speed 0.6 m/s by applying a force of 350 N. How
much work will be done during 0.5 h?
5. How much work is required to lift a 500 kg block12 m?
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6. A light plane travels a distance of 150 m along a runway before takeoff. Find the
work done by the plane engine if it is applying a force of 13500 N.
7. A railroad car is pulled through the distance of 960 m by a train that did 578 kJ of
work during this pull. How much force did the train supply?
8. A horse pulls a carriage by applying 450 N of force. Find the traveled distance if
the horse did 89 kJ of work.
9. A truck travels at a constant speed of 45 m/s. How much work did the truck
engine do during a 2 hour period if it supplied a force of 25 kN of force.
10. An airflow lifts a 3.6 kg bird 50 m up. How much work was done by the flow?
11. How much work must be done to lift a 4 m long chain 4 m off the floor by pulling
on one of its ends? Initially the chain was lying horizontally on the floor.
Kinetic Energy (KE)
1. A child does 12 J of work pushing his 3 Kg toy truck. With what velocity does the
toy move after the child is done pushing?
2. How much kinetic energy does an 80Kg man have while running at 1.5 m/s?
3. An object accelerates from 12m/s to 24 m/s in 3 s. What is its change in kinetic
energy?
4. A bullet is fired into a 12 kg block of wood. After the bullet stops in the block of
wood the block has 29 J of kinetic energy. At what speed is the block moving?
5. How much kinetic energy does a 4 Kg bird have while flying at 14m/s? How
much does it have at 18m/s?
6. How much work is required to stop an 8 kg medicine ball falling at 8m/s?
7. A 400 Kg car has 1.8 x 105 J of kinetic energy. How fast is it moving?
8. How fast is a 3 Kg toy car with 20 J of kinetic energy moving?
Gravitational Potential Energy (GPE)
1. A small, 3 Kg weight is moved from 5 m from the ground to 8 m. What is the
change in potential energy?
2. A Gravitational Potential Energy (GPE) Sensor attached to a 12 Kg ball changes
from 12 J to 22 J, by height change alone. What is the change in height?
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3. What is the gravitational potential energy of a 450 Kg car at the top of a 25 m
parking garage?
4. What is the change in gravitational potential energy of a 45 kg weight that is
moved from 2 m to 18 m one earth? What is it on the moon (g = 1.6 m/s2)?
5. A 24 g toy falls from 2 m to 1 m. What is the change in GPE?
6. If (on earth) an object falls 18 m and loses 36 J of GPE. What is the object’s
mass?
7. A 1 Kg object loses 20 J of GPE as it falls. How far does it fall?
8. An 80-Kg person falls 60 m off of a waterfall. What is her change in GPE?
Kinetic and Potential Energy
1. A 5 kg rock is raised 28 m above the ground level. What is the change in its
potential energy?
2. A 65 kg cart travels at constant speed of 4.6 m/s. What is its kinetic energy?
3. What is the potential energy of stretched spring, if the spring constant is 40 N/m
and the elongation is 5 cm?
4. A stone is thrown vertically up with a speed of 14 m/s, and at that moment it had
37 kJ of kinetic energy. What was the mass of the stone?
5. A 3.5 kg object gains 76 kJ of potential energy as it is lifted vertically. Find the
new height of the object?
6. A spring has a spring constant of 450 N/m. How much must this spring be
stretched to store 49 J of potential energy?
7. A projectile is fired vertically upward with an initial velocity of 190 m/s. Find the
maximum height of the projectile.
8. A spring gun is used to project a 500 g ball, in order to perform this experiment
the spring was initially compressed by 5 mm. Find the ball’s speed when it leaves
the gun, if the spring constant is 395 N/m.
9. An arrow is fired vertically upwards by a bow and reaches an altitude of 134 m.
Find the initial speed of the arrow on the ground level.
10. A roller coaster is released from the top of a track that is 125 m high. Find the
roller coaster speed when it reaches ground level.
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11. A 25 kg object is initially suspended from a rope that is 2.5 m long. The object is
pulled to the side and upwards, in the manner of a pendulum, and then released.
What is the maximum height from which it can be released, without breaking the
rope, if the rope can supply a maximum tension force of 550 N.
12. How much work must be done to accelerate an 8*105 kg train: a) from 10 m/s to
15 m/s; b) from 15 m/s to 20 m/s; c) to a stop an initial speed of 20 m/s?
13. An object is thrown vertically upward with an initial speed of 18 m/s. At what
height will its kinetic energy equal its potential energy?
14. A ball is thrown vertically downward from a height h above the ground level, after
rebounding it reaches a height of 2h. Find its initial velocity, assuming that the
collision was perfect elastic (mechanical energy is conserved).
Power
1. A heat engine does 23 kJ of work during 0.5 h. Find the power supplied by the
engine.
2. How much work is done by 15 kW engine during 3.5 h?
3. How long must a 400 W electrical engine work in order to produce 300 kJ of
work?
4. An elevator motor in a high-rise building can do 3500 kJ of work in 5 min. Find
the power developed by the motor.
5. A heat turbine can generate a maximum power of 250 MW. How much work can
the turbine do in 7.8 h?
6. How much time is required for a car engine to do 278 kJ of work, if its maximum
power is 95 kW?
7. How much time is required for a elevator to lift a 2000 kg load up 28 m from the
ground level, if the motor can produce 13 kW of power?
8. A 50 kW pump is used to pump up water from a mine that is 50 m deep. Find the
mass of water that can be lifted by the pump in 1.4 h.
9. Some scientists calculated that a whale can develop 150 kW of power when it is
swimming under the water surface at a constant speed 28 km/h. Find the
resistance force of the water exerted on the whale.
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10. A tractor travels at constant speed of 21.6 km/h. Find the power supplied by the
engine if it can supply a maximum force of 467 kN.
11. A 7.35 kW lathe can move an iron block at a constant speed by applying a force
of 5.56 kN. Find the speed of the block.
12. Find the power required by an airplane if it is to travel a distance of 100 m before
reaching its takeoff velocity of 25 m/s. Assume that the plane accelerates at a
constant rate and that its coefficient of air resistance is 0.02.
General Problems
1. A 255 N force is applied to a 46 kg box that is located on a flat horizontal surface.
The coefficient of kinetic friction between the box and the surface is 0.3.
a. Sketch a free-body diagram and show all the applied forces.
b. Find the acceleration of the box
c. How far the box will go in 10 s?
d. What will be the velocity at the end of this distance?
e. Find the kinetic energy after 10 s of traveling.
f. How much work is done during the first ten seconds by each of
the following; the applied force, friction force, normal force,
gravitational force and net force?
g. Compare the work done by the net force and the final kinetic
energy.
2. A worker pushes a 50 kg crate a distance of 7.5 m across a level floor. He
pushes it at a constant speed by applying a constant horizontal force. The
coefficient of kinetic friction between the crate and the floor is 0.15.
a. Find the magnitude of the applied force.
b. How much work did the worker do on the crate?
c. How much work did the friction force do on the crate?
d. How much work did the normal force do on the crate?
e. How much work did the gravitational force do on the crate?
f. What was the total work done on the crate?
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g. What was the change in the kinetic energy of the crate.
3. The diagram above shows a ballistic pendulum. A 10 g bullet is fired into the
suspended 2 kg block of wood and remains embedded inside it (a perfectly
inelastic collision). After the impact of the bullet, the block swings up to a
maximum height h. If the initial speed of the bullet was 35 m/s:
a. What was the momentum of the bullet before the collision?
b. What was the kinetic energy of the bullet before the collision?
c. What was the velocity of the bullet-block system just after the
collision?
d. What was the total kinetic energy of the bullet-block system after
the collision?
e. What is the maximum possible potential energy of the bullet-block
system when it reaches its maximum height?
f. What is the maximum possible height of the bullet-block system?
A
B
4. Two objects, A and B, with masses of 3.2 kg and 1.8 kg, move on a frictionless
horizontal surface. Object A moves to the right at a constant speed of 5.1 m/s
while object B moves to the right at a constant speed 1.4 m/s. They collide and
stick together (a perfectly inelastic collision).
a. Determine the total momentum of the system (both objects)
before the collision
b. Determine the total kinetic energy of the system before the
collision
c. Find the speed of the two objects after the collision
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d. Find the total kinetic energy of the system after the collision.
e. Is the kinetic energy of the system conserved? Explain.
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5. A small block, with a mass of 250 g, starts from rest at the top of the apparatus
shown above. It then slides without friction down the incline, around the loop and
then onto the final level section on the right. The maximum height of the incline is
80 cm, and the radius of the loop is 15 cm.
a. Find the initial potential energy of the block
b. Find the velocity the block at the bottom of the loop
c. Find the velocity of the block at the top of the loop.
d. What is the normal force on the block at the lowest point of the
loop?
e. What is the normal force on the block at the highest point of the
loop?
6. A 0.8 kg block is attached to the end of a spring whose spring constant is 85
N/m. The block is placed on a frictionless tabletop, given an initial displacement
of 3.5 cm and then released.
a. What type of energy did the block-spring system initially have?
b. Find the magnitude of this energy.
c. How does the total energy of the block-spring system change as
the block is pushed across the frictionless surface? Explain.
d. Find the maximum velocity of the block.
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7. A small cube, with a mass of 25 g, slides along a frictionless horizontal surface at
a constant speed of 18 m/s until it collides with, and sticks to, a large wooden 3.5
kg block. The large block is attached to the left end of a spring as shown above.
a. What is the momentum of the cube before the collision?
b. What is the kinetic energy of the cube before the collision?
c. Find the speed of the combined cube and block system just after
the collision.
d. Find the kinetic energy of the cube-block system just after the
collision.
e. What is the maximum potential energy that can be stored in the
spring due to this collision?
f. How far w the cube-block system move before it stops?
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8. A track consists of a frictionless arc XY, which is a quarter-circle of radius 0.5 m,
and a rough section YZ. Block A, whose mass is1.5 kg, is released from rest at
point X, slides down the curved section of the track, and collides instantaneously
and inelastically with identical block B at point Y. The two blocks move to the
right through the rough section of the track until they stop.
a. Determine the potential energy of block A at point X.
b. Determine the kinetic energy of block A at point Y, just before the
collision.
c. Find the speed of the two blocks just after the collision, at point Y.
d. Find the kinetic energy of the two blocks just after the collision, at
point Y.
e. How far will the two blocks travel on the rough section of the
track?
f. How much work will the friction force do during this time?
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30
Force (N)
15
10
20
30
Displacement (m)
9. An external horizontal force, F, is applied to a 2.5 kg toy car as it moves in a
straight line. The force varies with the car’s displacement as shown above. Use
the graph to answer the following questions.
a. How much work does the applied force do while the car does moves
the first 10 m?
b. Determine the kinetic energy of the car when it passes the 10 m point?
c. What is the velocity of the car when it passes the 10m point?
d. What is the total work done by the force in the process of displacing
the car the first 30 m?
e. What is the kinetic energy of the car when it is 30 m from the origin?
f. What is the velocity of the car when it is 30 m from the origin?
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10. A 2 kg object moves along a straight line. The net force acting on the object
varies with the object’s displacement as shown in the diagram above. The object
starts from rest at displacement x = 0 and time t = 0 and is travels a distance 20
m. Find the following.
a. The acceleration of the object when it has traveled 5 m.
b. The time taken for the object to be move the first 12 m.
c. The amount of work done by the net force in displacing the object
the first 12 m.
d. The speed of the object at a displacement of 12 m.
e. The speed of the object at a displacement 20 m.
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